Question Video: Finding Acceleration from a Velocity-Time Graph | Nagwa Question Video: Finding Acceleration from a Velocity-Time Graph | Nagwa

# Question Video: Finding Acceleration from a Velocity-Time Graph Mathematics • Third Year of Secondary School

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Below is a velocity–time graph of a marathon runner running down a racetrack over a period of 8 seconds. Find the acceleration of the runner during the first 2 seconds. Find the distance covered by the runner over the 8-second period.

03:06

### Video Transcript

Below is a velocity–time graph of a marathon runner running down a racetrack over a period of eight seconds. Find the acceleration of the runner during the first two seconds. Find the distance covered by the runner over the eight-second period.

When trying to work out the acceleration from a velocity–time graph, we need to work out the slope or gradient of the graph. The acceleration is equal to the change in velocity divided by the change in the time. The acceleration of the runner in the first four seconds is constant. Therefore, this will be the acceleration during the first two seconds. We see here that we have a change in velocity of six meters per second, from three to nine meters per second. We have a change in time of four seconds. This means that the acceleration is equal to six divided by four. This is sometimes known as the rise over the run.

The fraction six over four simplifies to three over two by dividing the numerator and denominator by two. This is equal to 1.5. As the velocity was measured in meters per second and the time in seconds, the acceleration will be measured in meters per second per second, otherwise known as meters per second squared. The acceleration of the runner during the first two seconds is 1.5 meters per second per second.

We can calculate the distance covered by the runner by working out the area between the graph and the 𝑥-axis. This can be split into two shapes: a trapezoid, or trapezium, and a rectangle. To calculate the area of a trapezoid, we add the lengths of the parallel sides and divide by two. We then multiply this by the perpendicular height between them. The parallel sides are of lengths three and nine. And the height or distance between them is four. Three plus nine is equal to 12. Halving this gives us six. And multiplying by four gives us 24 meters.

The area of any rectangle is equal to its length multiplied by its width. In this case, nine multiplied by four is equal to 36. The distance covered in this part of the journey is 36 meters. 24 plus 36 is equal to 60. Therefore, the distance covered by the runner over the eight-second period is 60 meters.

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